Prove that $‎f(x+h)+f(x-h)-2f(x)‎\leq0 ‎\;\;‎;\; ‎h‎>0‎$‎

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a concave function, then prove that $$‎‎f(x+h)+f(x-h)-2f(x)‎\leq0\,;‎\quad ‎h‎>0‎‎‎$$‎‎

I tried to prove this by concave definition as follows:

If $f$ is concave function then

$$‎‎f(‎\lambda ‎x+(1-‎\lambda) ‎y‎‎)‎\geq‎‎\lambda‎ f(x)+(1-‎\lambda‎)f(y)\,;‎\quad0‎<‎\lambda‎‎<1‎‎‎$$‎‎ After that I applied the above for $f(x+h)$ and $f(x-h)$ as follows:

$$f(x+h)=f\left(\frac{1}{2}(2x)+\frac{1}{2}(2h)\right)\geq\frac{1}{2}f(2x)+\frac{1}{2}f(2h)\\ f(x-h)=f\left(\frac{1}{2}(2x)+\frac{1}{2}(-2h)\right)\geq\frac{1}{2}f(2x)+\frac{1}{2}f(-2h)$$

Finally I combined the two above inequality but I could not arrive at the result.
Any idea can help me? Thanks.

Answer 1

Because $$\frac{f(x+h)+f(x-h)}{2}\leq f\left(\frac{x+h+x-h}{2}\right)=f(x).$$

Answer 2

Hint: Consider taking $x=x+h$ and $y=x-h$ and $\lambda$ as assumed, in $$‎‎f(‎\lambda ‎x+(1-‎\lambda) ‎y‎‎)‎\geq‎‎\lambda‎ f(x)+(1-‎\lambda‎)f(y)\;\;;‎0‎<‎\lambda‎‎<1‎‎‎$$

Important to note both $x$ and $y$ must be in the same interval in consideration.

Answer 3

Hint: let $\lambda=1/2$ and compute $2f(\lambda(x+h) + (1-\lambda)(x-h))$.

Answer 4

I suspect there's an insightful understanding that goes with your question as it's so similar to the definition of a gradient from first principles; $$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

So maybe a small manipulation such as; $$f(x+h)-f(x-h)-2f(x)=f(x+h)-f(x)-(f(x)-f(x-h))$$ and then think about

$$\lim_{h \to 0} \frac{f(x+h)-f(x-h)-2f(x)}{h}$$ $$=\lim_{h \to 0}\big( \frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}\big)$$

Is it obvious this will be concave ?

I suspect it's the $h \gt 0$ is the crucial piece of information that means it is concave (whereas if $h \lt 0$ it would be convex)

So there may be some playing around with "tending to zero from above" and "tending to zero from below" to get this technically correct.

Answer 5

Hint

Define $u=x+h$ and $v=x-h$. Use the definition of concavity for $u,v$ and for appropriate value of $\lambda$.